Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int iinf = 1e9 + 7;
const long long linf = 1ll << 60;
const double dinf = 1e10;
inline long long read() {
  bool f = 0;
  long long x = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
  return x;
}
inline void scf(int &x) {
  bool f = 0;
  x = 0;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') {
    f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
  return;
}
void scf(int &x, int &y) {
  scf(x);
  return scf(y);
}
void scf(int &x, int &y, int &z) {
  scf(x);
  scf(y);
  return scf(z);
}
const int N = 1e5 + 100;
const int lgN = 18;
int n, m, q;
vector<int> g[N], has[N];
int dep[N], par[N][lgN];
struct min_p {
  int len, p[10];
  min_p operator+(const min_p &a) const {
    min_p ret;
    ret.len = 0;
    int i = 0, j = 0;
    while (i < len && j < a.len) {
      const int &x = p[i], &y = a.p[j];
      if (x < y)
        ret.p[ret.len++] = x, i++;
      else
        ret.p[ret.len++] = y, j++;
      if (ret.len == 10) return ret;
    }
    while (i < len && ret.len < 10) ret.p[ret.len++] = p[i++];
    while (j < a.len && ret.len < 10) ret.p[ret.len++] = a.p[j++];
    return ret;
  }
  inline void out(int siz = 10) const {
    if (len < siz) siz = len;
    printf("%d", siz);
    for (int i = 0; i < (siz); i++) printf(" %d", p[i]);
    putchar('\n');
    return;
  }
} p[N][lgN];
inline void dfs(int u = 1, int f = 0) {
  par[u][0] = f;
  dep[u] = dep[f] + 1;
  for (int i = 0; i < (lgN - 1); i++) par[u][i + 1] = par[par[u][i]][i];
  min_p &t = p[u][0];
  t.len = 0;
  for (auto v : has[u]) {
    t.p[t.len++] = v;
    if (t.len == 10) break;
  }
  for (int i = 0; i < (lgN - 1); i++) p[u][i + 1] = p[u][i] + p[par[u][i]][i];
  for (auto v : g[u])
    if (v != f) dfs(v, u);
  return;
}
inline void LCA(int u, int v, const int &k) {
  if (dep[u] < dep[v]) u ^= v, v ^= u, u ^= v;
  min_p ret;
  ret.len = 0;
  int dlt = dep[u] - dep[v];
  for (int i = 0; i < (lgN); i++)
    if ((dlt >> i) & 1) ret = ret + p[u][i], u = par[u][i];
  if (u == v) {
    ret = ret + p[u][0];
    return ret.out(k);
  }
  for (int i = (lgN - 1); i >= (0); i--)
    if (par[u][i] != par[v][i])
      ret = ret + p[u][i] + p[v][i], u = par[u][i], v = par[v][i];
  ret = ret + p[u][1] + p[v][0];
  return ret.out(k);
}
int main() {
  scf(n, m, q);
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    scf(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = (1); i <= (m); i++) {
    int u;
    scf(u);
    has[u].push_back(i);
  }
  dfs();
  while (q--) {
    int u, v, k;
    scf(u, v, k);
    LCA(u, v, k);
  }
  return 0;
}